Chapter 9.4, Problem 52SR

To determine

To find: Write an equation for the circle that satisfies each set of conditions.

Answer to Problem 52SR

The equation for the circle is ${(x-4)}^2+{(y-2)}^2=4$ .

Explanation of Solution

Given information:

The given centre is,

  $\left(4,2\right)$

Tangent to $x-axis$

The standard form of a circle is

  ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$

There are $\left(h,k\right)$ is the centre point and $r$ is the radius of the circle

The expression is simplified as,

  ${\left(x-4\right)}^2+{\left(y-2\right)}^2=r^2\mathrm{....}\left(1\right)$

The circle is tangent to the $\text{x-axis}$ therefore the radius is $2$

Simplify further to substitute the value ofradius $r=2,$ in the equation $\mathrm{...}\left(1\right)$ ,

  $\begin{array}{l}{\left(x-4\right)}^2+{\left(y-2\right)}^2=r^2\\ {\left(x-4\right)}^2+{\left(y-2\right)}^2=2^2\\ {\left(x-4\right)}^2+{\left(y-2\right)}^2=4\end{array}$

Thus, the equation for the circle is ${(x-4)}^2+{(y-2)}^2=4$ .